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olga2289 [7]
3 years ago
11

The slope-intercept form of the equation of a line that passes through point (–3, 8) is y = –2/3x + 6. What is the point-slope f

orm of the equation for this line? y – 3 = –2/3(x + 8) y + 3 = –2/3(x – 8) y + 8 = –2/3(x – 3) y – 8 = –2/3(x + 3)
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
8 0
I think the correct answer would be the last option. The correct point slope form would be <span>y – 8 = –2/3(x + 3). To write  this, we only follow the general formula y - y1 = m(x -x1) where m is the slope which is -2/3 and x1 and y1 is the point given above. We do as follows:</span>

y - y1 = m(x-x1)
y -  8 = -2/3 (x + 3)
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Find the exact length of the curve. 36y2 = (x2 − 4)3, 5 ≤ x ≤ 9, y ≥ 0
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We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.

We are given: 36 y^{2} =( x^{2} -4)^3
We divide by 36 and take the root of both sides to obtain: y = \sqrt{ \frac{( x^{2} -4)^3}{36} }

Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain: y =  \frac{( x^{2} -4)^{3/2}}{6} }=( \frac{1}{6} )(x^{2} -4)^{3/2}}

Let's leave that for the moment and look at the formula for arc length. The formula is L= \int\limits^c_d {ds} where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.

Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows: ds= \sqrt{1+( \frac{dy}{dx})^2 } dx

As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.

As you can see from the formula we need to find dy/dx and square it. Let's do that now.

We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here x^2-4). More formally, we can let u=x^{2} -4 and then consider the derivative of u^{3/2}du. Either way, we obtain,

\frac{dy}{dx}=( \frac{1}{6})( x^{2} -4)^{1/2}(2x)=( \frac{x}{2})( x^{2} -4)^{1/2}

Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
( \frac{dy}{dx}^2)=( \frac{x^2}{4})( x^{2} -4)= \frac{x^4-4 x^{2} }{4}

This means that in our case:
ds= \sqrt{1+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{4}{4}+\frac{x^4-4 x^{2} }{4}} dx
ds= \sqrt{\frac{x^4-4 x^{2}+4 }{4}} dx
ds= \sqrt{\frac{( x^{2} -2)^2 }{4}} dx
ds=  \frac{x^2-2}{2}dx =( \frac{1}{2} x^{2} -1)dx

Recall, the formula for arc length: L= \int\limits^c_d {ds}
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:

L= \int\limits^9_5 { \frac{1}{2} x^{2} -1 } \, dx = (\frac{1}{2}) ( \frac{x^3}{3}) -x evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.

That is, [(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}


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